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In topology, the “shrinking lemma” (lemma below) states that on a normal topological space the patches of every locally finite cover may be replaced by smaller patches which still cover the space, but such that their topological closures are contained in the original patches.
If there is more than a countable set of elements in the original cover, then this conclusion requires excluded middle and Zorn's lemma, hence the axiom of choice.
The shrinking lemma is needed in the proof that paracompact Hausdorff spaces equivalently admit subordinate partitions of unity.
(shrinking lemma for locally finite covers)
Assuming the axiom of choice then:
Let $X$ be a topological space which is normal and let $\{U_i \subset X\}_{i \in I}$ be a locally finite open cover.
Then there exists another open cover $\{V_i \subset X\}_{i \in I}$ such that the topological closure $Cl(V_i)$ of its elements is contained in the original patches:
We now prove this in increasing generality, first for binary open covers (lemma below), then for finite covers (lemma ), then for locally finite countable covers (lemma ), and finally for general locally finite covers (lemma , proof below). It is only the last statement that needs the axiom of choice.
(shrinking lemma for binary covers)
Let $(X,\tau)$ be a normal topological space and let $\{U_i \subset X\}_{i \in \{1,2\}}$ an open cover by two open subsets.
Then there exists an open set $V_1 \subset X$ whose topological closure is contained in $U_1$
and such that $\{V_1,U_2\}$ is still an open cover of $X$.
Since $U_1 \cup U_2 = X$ it follows (by de Morgan's law) that their complements $X \backslash U_i$ are disjoint closed subsets. Hence by normality of $(X,\tau)$ there exist disjoint open subsets
By their disjointness, we have the following inclusions:
In particular, since $X \backslash V_2$ is closed, this means that $Cl(V_1) \subset X \backslash V_2$.
Hence it only remains to observe that $V_1 \cup U_2 = X$, by definition of $V_1$.
(shrinking lemma for finite covers)
Let $(X,\tau)$ be a normal topological space, and let $\{U_i \subset X\}_{i \in \{1, \cdots, n\}}$ be an open cover with a finite number $n \in \mathbb{N}$ of patches. Then there exists another open cover $\{V_i \subset X\}_{i \in I}$ such that $Cl(V_i) \subset U_i$ for all $i \in I$.
By induction using lemma .
To begin with, consider $\{ U_1, \underoverset{i = 2}{n}{\cup} U_i\}$. This is a binary open cover, and hence lemma gives an open subset $V_1 \subset X$ with $V_1 \subset Cl(V_1) \subset U_1$ such that $\{V_1, \underoverset{i = 2}{n}{\cup} U_i\}$ is still an open cover, and accordingly so is
Similarly we next find an open subset $V_2 \subset X$ with $V_2 \subset Cl(V_2) \subset U_2$ and such that
is an open cover. After $n$ such steps we are left with an open cover $\{V_i \subset X\}_{i \in \{1, \cdots, n\}}$ as required.
Beware that the induction in lemma does not give the statement for infinite countable covers. The issue is that it is not guaranteed that $\underset{i \in \mathbb{N}}{\cup} V_i$ is a cover.
And in fact, assuming the axiom of choice, then there exists a counter-example of a countable cover on a normal spaces for which the shrinking lemma fails (a Dowker space due to Beslagic 85).
This issue is evaded if we consider locally finite covers:
(shrinking lemma for locally finite countable covers)
Let $(X,\tau)$ be a normal topological space and $\{U_i \subset X\}_{i \in \mathbb{N}}$ a locally finite countable cover. Then there exists open subsets $V_i \subset X$ for $i \in \mathbb{N}$ such that $V_i \subset Cl(V_i) \subset U_i$ and such that $\{V_i \subset X\}_{i \in \mathbb{N}}$ is still a cover.
As in the proof of lemma , there exist $V_i$ for $i \in \mathbb{N}$ such that $V_i \subset Cl(V_i) \subset U_i$ and such that for every finite number, hence every $n \in \mathbb{N}$, then
Now the extra assumption that $\{U_i \subset X\}_{i \in I}$ is locally finite implies that every $x \in X$ is contained in only finitely many of the $U_i$, hence that for every $x \in X$ there exists $n_x \in \mathbb{N}$ such that
This implies that for every $x$ then
hence that $\{V_i \subset X\}_{i \in \mathbb{N}}$ is indeed a cover of $X$.
We now invoke Zorn's lemma to generalize the shrinking lemma for finitely many patches (lemma ) to arbitrary sets of patches:
of the general shrinking lemma
Let $\{U_i \subset X\}_{i \in I}$ be the given locally finite cover of the normal space $(X,\tau)$. Consider the set $S$ of pairs $(J, \mathcal{V})$ consisting of
a subset $J \subset I$;
an $I$-indexed set of open subsets $\mathcal{V} = \{V_i \subset X\}_{i \in I}$
with the property that
$(i \in J \subset I) \Rightarrow ( Cl(V_i) \subset U_i )$;
$(i \in I \backslash J) \Rightarrow ( V_i = U_i )$.
$\{V_i \subset X\}_{i \in I}$ is an open cover of $X$.
Equip the set $S$ with a partial order by setting
By definition, an element of $S$ with $J = I$ is an open cover of the required form.
We claim now that a maximal element $(J, \mathcal{V})$ of $(S,\leq)$ has $J = I$.
For assume on the contrary that there were $i \in I \backslash J$. Then we could apply the construction in lemma to replace that single $V_i$ with a smaller open subset $V'_i$ to obtain $\mathcal{V}'$ such that $Cl(V'_i) \subset V_i$ and such that $\mathcal{V}'$ is still an open cover. But that would mean that $(J,\mathcal{V}) \lt (J \cup \{i\}, \mathcal{V}')$, contradicting the assumption that $(J,\mathcal{V})$ is maximal. This proves by contradiction that a maximal element of $(S,\leq)$ has $J = I$ and hence is an open cover as required.
We are reduced now to showing that a maximal element of $(S,\leq)$ exists. To achieve this we invoke Zorn's lemma. Hence we have to check that every chain in $(S,\leq)$, hence every totally ordered subset has an upper bound.
So let $T \subset S$ be a totally ordered subset. Consider the union of all the index sets appearing in the pairs in this subset:
Now define open subsets $W_i$ for $i \in K$ picking any $(J,\mathcal{V})$ in $T$ with $i \in J$ and setting
This is independent of the choice of $(J,\mathcal{V})$, hence well defined, by the assumption that $(T,\leq)$ is totally ordered.
Moreover, for $i \in I\backslash K$ define
We claim now that $\{W_i \subset X\}_{i \in I}$ thus defined is a cover of $X$. Take an arbitrary point $x \in X$. If $x \in U_i$ for some $i \notin K$, we have $U_i = W_i$ and therefore $x$ is in $\underset{i \in I}{\cup} W_i$. Otherwise, combining with the assumption that $\{U_i \subset X\}_{i \in I}$ is locally finite, the set $J_x$ of indices $i \in I$ such that $x \in U_i$ is finite and $J_x \subset K$. Since $(T,\leq)$ is a total order, it must contain an element $(J, \mathcal{V})$ such that $J_x \subset J$. And since that $\mathcal{V}$ is a cover and $x$ cannot belong to any $U_i$ with $i$ outside of $J_x$, it must be that $x \in \underset{i \in J_x}{\cup} V_i \subset \underset{i \in J}{\cup} V_i$, and hence $x$ is in $\underset{i \in I}{\cup} W_i$.
This shows that $(K,\mathcal{W})$ is indeed an element of $S$. It is clear by construction that it is an upper bound for $(T ,\leq )$. Hence we have shown that every chain in $(S,\leq)$ has an upper bound, and so Zorn’s lemma implies the claim.
The above account follows
The example (a Dowker space) of a normal space with a (not locally-finite) countable cover to which the shrinking lemma does not apply is given in
Last revised on June 24, 2017 at 00:55:58. See the history of this page for a list of all contributions to it.